Question

How do you evaluate the limit `tan(3x)/(3tan2x)` as x approaches `0`?

Answer 1

`Lt_(x->0)(tan(3x))/(3tan2x)=1/2`

Explanation:

Let us first find `Lt_(x->0)tanx/x`

`Lt_(x->0)tanx/x=Lt_(x->0)(sinx)/(xcosx)`

= `Lt_(x->0)(sinx)/x xx Lt_(x->0)1/cosx`

= `1xx1=1`

Hence `Lt_(x->0)(tan(3x))/(3tan2x)`

= `Lt_(x->0)((tan(3x))/(3x))/(3(tan2x)/(2x))xx(3x)/(2x)`

= `(Lt_(3x->0)((tan3x)/(3x)))/(3Lt_(2x->0)((tan2x)/(2x)))xx3/2`

= `1/(3xx1)xx3/2=1/2`

Answer 2

`lim_(xrarr0)tan(3x)/(3tan(2x))=1/2`

Explanation:

Use `tan(a+b)=(tan(a)+tan(b))/(1-tan(a)tan(b))` to rewrite `tan(3x)` as `tan(2x+x)`:

`lim_(xrarr0)tan(3x)/(3tan(2x))=lim_(xrarr0)((tan(2x)+tan(x))/(1-tan(2x)tan(x)))/(3tan(2x))`

`=lim_(xrarr0)(tan(2x)+tan(x))/(3tan(2x)(1-tan(x)tan(2x)))`

Now using `tan(2x)=(2tan(x))/(1-tan^2(x))`:

`=lim_(xrarr0)((2tan(x))/(1-tan^2(x))+tan(x))/(3(2tan(x))/(1-tan^2(x))(1-tan(x)(2tan(x))/(1-tan^2(x))))`

Multiplying through by `1-tan^2(x)`:

`=lim_(xrarr0)(2tan(x)+tan(x)(1-tan^2(x)))/(6tan(x)(1-(2tan^2(x))/(1-tan^2(x)))`

Factoring `tan(x)` then cancelling:

`=lim_(xrarr0)(tan(x)(2+1-tan^2(x)))/(6tan(x)(1-(2tan^2(x))/(1-tan^2(x)))`

`=lim_(xrarr0)(2+1-tan^2(x))/(6(1-(2tan^2(x))/(1-tan^2(x)))`

Now we can plug in `x=0` and evaluate the limit without a problem:

`=(2+1-0)/(6(1-(2(0))/(1-0)))=3/6=1/2`